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TPS2420: CT discharge current seems incorrect.

Cameron Chapman
Prodigy 60 points
Part Number: TPS2420

CT desc says 5uA discharges CT after a fault, but block diagram says 1.25uA and TSD equation calculates to a bout 1.2uA. Which is it? Could the retry time be sped up by switching a resistor in parallel with CT? Are there any limits to how much I could speed it up? 

  • 0
    TI__Guru 57737 points

    Hi Cameron,

    As you mentioned, the datasheet is a bit confusing and needs modification. For your understanding below is what you can consider. 

    • The fault timer starts when IOUT exceeds IFAULT or when SOA protection mode is active, charging the CT capacitor with 35 μA from GND towards an upper threshold of 1.4 V.
    • If the capacitor reaches the upper threshold, the internal pass MOSFET is turned off.
    • If the device is in Latch mode, the MOSFET remains off until EN is cycled.
    • If the device is in retry mode , the capacitor will discharge at 1.4 μA to 0.2 V and then re-enable the pass MOSFET.
      • The 1.4uA is typical value and this can vary from 1uA to 1.8uA. In Tsd equation you can consider 1.4uA.
    • If the upper threshold (1.4 V) is not crossed, the capacitor will discharge at 33 μA to 0.2 V and then to 0 V at 1.4 μA.
    • When the device is disabled, CT is pulled to GND through a 100-kΩ resistor.

    The reason for the discharge current in retry mode to be less is to give the IC time to cool down. When the device is operated in conditions like over Current and SOA regions (during start up) the device internal temperature increases due to the power dissipation in the internal FET. We have to give enough time for the FET to cool down before we put it under similar condition again or else the device can shut down due to thermal limit. 

  • 0 in reply to Praveen GD
    Prodigy 60 points

    Thank you Praveen. It seems like the 1.25uA value on page 5 and the 5uA value on page 7 are wrong and should be corrected to 1.4uA. Do you agree? 

    For our use, we are usually operating with low SS current and lower temperatures. So, we would be ok with a shorter cooldown time, even if it risked possibly shutting off due to thermal limit to cool further. Would adding a resistor switched by !FLT in parallel with CT create any other issues? 

  • 0 in reply to Cameron Chapman
    TI__Guru 57737 points
    Hi Cameron,

    Yes, for CT Sinking Current you can consider 1.4V as typical and the value can vary between 1V to 1.8V. The 5uA mentioned on page 7 needs to be fixed.
    I don't see a problem connecting a resistor between CT and FLT/ as long as you don't have a pullup resistor on FLT/. We have not tested this circuit so we recommend you to thoroughly test the circuit at all conditions before going ahead with it.
  • 0 in reply to Praveen GD
    Prodigy 60 points

    Thanks Praveen.

    I meant that I would use FLT/ to switch on a transistor, that would then connect a resistor across CT to help it drain faster. Other than reducing cool down time, do you predict any other issues doing that? 

  • 0 in reply to Cameron Chapman
    TI__Guru 57737 points
    Hi Cameron,

    I don't see any obvious issues other than the Thermal Shutdown issue. As mentioned earlier please test the circuit thoroughly before implementing ni your final design.